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Chapter 9: Problem 43

A frictionless pulley has the shape of a uniform solid disk of mass 2.50 kg and radius 20.0 cm. A 1.50-kg stone is attached to a very light wire that is wrapped around the rim of the pulley (\(\textbf{Fig. E9.43}\)), and the system is released from rest. (a) How far must the stone fall so that the pulley has 4.50 J of kinetic energy? (b) What percent of the total kinetic energy does the pulley have?

Short Answer

Expert verified
(a) The stone must fall 0.167 m. (b) The pulley has 45.6% of the total kinetic energy.

Step by step solution

01

Identify Known Quantities

The mass of the pulley, \(m_p = 2.50 \text{ kg}\). The radius of the pulley, \(r = 0.20 \text{ m}\). The mass of the stone, \(m_s = 1.50 \text{ kg}\). The initial kinetic energy is \(0\) J because the system starts from rest.
02

Understand Total Kinetic Energy of the System

The total kinetic energy of the system is given as \(4.50 \text{ J}\). This combines the translational kinetic energy of the stone and the rotational kinetic energy of the pulley.
03

Expression for Kinetic Energy

The total kinetic energy \(K\) for the system is the sum of the translational kinetic energy of the stone and the rotational kinetic energy of the pulley:\[ K = \frac{1}{2} m_s v^2 + \frac{1}{2} I \omega^2 \] where \(v\) is the speed of the stone and \(\omega\) is the angular velocity of the pulley.
04

Relate Angular and Linear Velocity

The angular velocity \(\omega\) is related to the linear velocity \(v\) by the equation \(\omega = \frac{v}{r}\). Substitute \(\omega\) into the expression for rotational kinetic energy:\[ \frac{1}{2} I \omega^2 = \frac{1}{2} \left( \frac{1}{2} m_p r^2 \right) \left( \frac{v}{r} \right)^2 = \frac{1}{4} m_p v^2 \]
05

Substitute Back to Total Kinetic Energy Equation

Substitute \(\frac{1}{4} m_p v^2\) for the rotational kinetic energy into the total kinetic energy equation:\[ 4.50 = \frac{1}{2} m_s v^2 + \frac{1}{4} m_p v^2 \]Combine terms:\[ = \left( \frac{1}{2} m_s + \frac{1}{4} m_p \right) v^2 \]
06

Solve for Velocity \(v\)

Solve for \(v\):\[ 4.50 = \left( \frac{1}{2} \times 1.50 + \frac{1}{4} \times 2.50 \right) v^2 \]\[ 4.50 = (0.75 + 0.625) v^2 \]\[ 4.50 = 1.375 v^2 \]\[ v^2 = \frac{4.50}{1.375} \]\[ v = \sqrt{\frac{4.50}{1.375}} \approx 1.81 \text{ m/s}\]
07

Calculate Distance Fallen

Use the kinematic equation relating velocity, acceleration due to gravity \(g\), and distance \(d\):\[ v^2 = 2gd \]Solving for \(d\):\[ d = \frac{v^2}{2g} = \frac{1.81^2}{2 \times 9.8} \approx 0.167 \text{ m}\]
08

Calculate Percentage of Total Kinetic Energy for the Pulley

The kinetic energy of the pulley \(K_p\) is given by:\[ K_p = \frac{1}{4} m_p v^2 \]Substitute the known values:\[ K_p = \frac{1}{4} \times 2.50 \times 1.81^2 \approx 2.05 \text{ J}\]The percentage of the pulley's energy to total energy:\[ \text{Percentage} = \left( \frac{2.05}{4.50} \right) \times 100\% \approx 45.6\% \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Rotational Motion
Rotational motion is a concept you encounter in systems where objects spin around an axis. In this exercise, the pulley represents a body that undergoes rotational motion. Unlike linear motion, rotational motion concerns quantities like angular velocity \( \omega \), which describes how fast something spins.
  • **Angular velocity** \( \omega \) is related to linear velocity \( v \) through the radius \( r \) of the pulley: \( \omega = \frac{v}{r} \).
  • The pulley's mass and radius are crucial to determining its ability to rotate since they influence its moment of inertia \( I \).
Understanding rotational motion helps in analyzing systems with rotating parts, like this one where a pulley determines how a stone moves downwards. This kind of motion is crucial in real-world applications ranging from engines to amusement park rides.
Kinetic Energy
Kinetic energy is the energy an object has due to its motion. In this exercise, we focus on two forms: translational kinetic energy and rotational kinetic energy.
  • **Translational kinetic energy** for the falling stone is given by: \[ KE_{trans} = \frac{1}{2} m_s v^2 \]
  • **Rotational kinetic energy** for the pulley is expressed as: \[ KE_{rot} = \frac{1}{2} I \omega^2 \]
For the pulley, we convert \( \omega \) using \( \omega = \frac{v}{r} \), resulting in a term where the rotational energy depends on \( v^2 \) and other parameters of the pulley, like mass \( m_p \) and radius \( r \). Understanding how these two energies combine is critical; the total kinetic energy of this system is the sum: \[ K = \frac{1}{2} m_s v^2 + \frac{1}{4} m_p v^2 \].This equation allows us to solve for the speed \( v \) and analyze the stone's motion as well as the energy dynamics in a frictionless system.
Frictionless Systems
In a frictionless system, energy transformations occur without the loss of energy due to heat or resistance. Theoretically, these systems allow us to focus solely on conversion between potential and kinetic energy.
  • Without friction, the only forces acting are due to gravity, causing objects like the stone to fall.
  • This exercise has no energy loss due to friction, which simplifies calculations and ensures that all potential energy converts to kinetic energy.
The benefit of considering such idealized systems is the ability to predict outcomes with greater accuracy in an educational sense. These principles apply, for example, in systems designed to minimize energy loss in engineering contexts, enhancing efficiency.

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Most popular questions from this chapter

A uniform sphere with mass 28.0 kg and radius 0.380 m is rotating at constant angular velocity about a stationary axis that lies along a diameter of the sphere. If the kinetic energy of the sphere is 236 J, what is the tangential velocity of a point on the rim of the sphere?

A thin, light wire is wrapped around the rim of a wheel as shown in Fig. E9.45. The wheel rotates about a stationary horizontal axle that passes through the center of the wheel. The wheel has radius 0.180 m and moment of inertia for rotation about the axle of \(I =\) 0.480 kg \(\cdot\) m\(^2\). A small block with mass 0.340 kg is suspended from the free end of the wire. When the system is released from rest, the block descends with constant acceleration. The bearings in the wheel at the axle are rusty, so friction there does \(-\)9.00 J of work as the block descends 3.00 m. What is the magnitude of the angular velocity of the wheel after the block has descended 3.00 m?

The eel is observed to spin at 14 spins per second clockwise, and 10 seconds later it is observed to spin at 8 spins per second counterclockwise. What is the magnitude of the eel's average angular acceleration during this time? (a) 6/10 rad/s\(^2\); (b) 6\(\pi\)/10 rad/s\(^2\); (c) 12\(\pi\)/10 rad/s\(^2\); (d) 44\(\pi\)/10 rad/s\(^2\).

A circular saw blade with radius 0.120 m starts from rest and turns in a vertical plane with a constant angular acceleration of 2.00 rev/s\(^2\). After the blade has turned through 155 rev, a small piece of the blade breaks loose from the top of the blade. After the piece breaks loose, it travels with a velocity that is initially horizontal and equal to the tangential velocity of the rim of the blade. The piece travels a vertical distance of 0.820 m to the floor. How far does the piece travel horizontally, from where it broke off the blade until it strikes the floor?

A uniform bar has two small balls glued to its ends. The bar is 2.00 m long and has mass 4.00 kg, while the balls each have mass 0.300 kg and can be treated as point masses. Find the moment of inertia of this combination about an axis (a) perpendicular to the bar through its center; (b) perpendicular to the bar through one of the balls; (c) parallel to the bar through both balls; and (d) parallel to the bar and 0.500 m from it.
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